I Differentiable manifolds over fields other than R, C

WWGD

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[Moderator's note: Spin-off from another thread.]

Summary: How do you define a derivative on a manifold with no metric?

I was reading about differentiable manifolds on wikipedia, and in the definition it never specifies that the differentiable manifold has a metric on it. I understand that you can set up limits of functions in topological spaces without a metric being defined, but my understanding of derivatives suggests that you need a metric in both the domain and the codomain, in order to come up with a rate of change which you are finding the limit of. Is there a more general definition of the derivative that is being used here?
You need the structure of a topological vector field K with 0 as a limit point of K-{0}. The TVF structure allows the addition and quotient expression to make sense; you need 0 as a limit point to define the limit as h-->0 and the topology to speak of convergence and a limit.

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fresh_42

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Or to say it short: $\varphi\, : \,M\longrightarrow N \Longrightarrow D\varphi\, : \,TM\longrightarrow TN$ and the differentiation takes place in the tangent bundle, over a real or complex vector space.

Math_QED

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For a differentiable manifold you don't need the charts to be differentiable (smooth). You just require that each chart is a homeomorphism of an open subset of your topological space onto an open subset of an Euclidean space $\mathbb{R}^n$. The differentiability kicks in when we talk about compatibility of charts. If $\phi, \psi$ are two chart, we want that $\phi \circ \psi^{-1}$ is a smooth map on their common domain.

Or maybe I misunderstand the question and you ask about what a differentiable map between two (smooth) manifolds is? What do you mean with a derivative on a manifold? You can only take the derivative of a map.

WWGD

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Or to say it short: $\varphi\, : \,M\longrightarrow N \Longrightarrow D\varphi\, : \,TM\longrightarrow TN$ and the differentiation takes place in the tangent bundle, over a real or complex vector space.
I think you may be able to do it over the p-adics, but I have no idea how/what that would be like.

fresh_42

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I think you may be able to do it over the p-adics, but I have no idea how/what that would be like.
A non-Archimedean metric? Would be strange, but I have a bit of the suspicion that nobody actually has an idea of the p-adic world. On the other hand they play a role in my book about quadratic forms, and manifolds are not far away, at least orthogonal groups. I recently asked about the possibility to vary the field in a GUT instead of the symmetry groups. I only received a big no-no as answers, but this might have more to do with mainstream than with necessity. Why should it be impossible to build the standard model on orthogonal groups over p-adic vector spaces, possibly with a large p? But as long as nobody proves a version of Noether over p-adic fields, it is unlikely that anybody considers such an approach. They are too deeply caught in the superworld and real eigenvalues as observables.

WWGD

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By metric do you mean a distance metric or metric tensor? I don't see the need to use a distance metric to define the limit quotient; just need to define addition, quotient ( multiplicative inverse) and limits, and having {0} as a limit point of the complement.

fresh_42

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There is more to a manifold than differentiability. But even differentiability needs to have nearby defined! Sure, this can be done by open neighborhoods and thus by a p-adic topology. I only assume that a non-Archimedean distance will have strange consequences for the analysis on such a manifold.

If you think about it, then the entire (physical) world is build on the concept of invariance of a quadratic form. In GR it is $(-1,1,1,1)$, and in SM it are the unitary groups. Isn't that strange? You only need to be able to distinguish left and right, forward and backward, i.e. $2\neq 0$. The rest is orthogonality.

WWGD

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Ah, yes, most defs of manifold imply metrizability. I think normal+ 2nd countable does it by, e.g.. Urysohn. I was thinking of differentiability as a stand-alone.

WWGD

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There is more to a manifold than differentiability. But even differentiability needs to have nearby defined! Sure, this can be done by open neighborhoods and thus by a p-adic topology. I only assume that a non-Archimedean distance will have strange consequences for the analysis on such a manifold.

If you think about it, then the entire (physical) world is build on the concept of invariance of a quadratic form. In GR it is $(-1,1,1,1)$, and in SM it are the unitary groups. Isn't that strange? You only need to be able to distinguish left and right, forward and backward, i.e. $2\neq 0$. The rest is orthogonality.
Edit: But these are not standard metrics since they are not Real- valued. In the non-standard Reals,e.g., if a is a non-standard number, d(a,0)=a is non-Real.

fresh_42

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But these are not standard metrics since they are not Real- valued. In the non-standard Reals,e.g., if a is a non-standard number, d(a,0)=a is non-Real.
But this is another issue and has nothing to do with p-adic completions. The most abstract concept to do analysis is probably measure theory. However, I have no idea how measure theory works over the p-adics.

The Minkowski metric or the unitarity of symmetry groups are all special versions of orthogonality, or better invariances of a quadratic form. My book about quadratic forms (O'Meara) has not only p-adics as examples, it also deals with Clifford algebras, spinors, and as mentioned, orthogonal groups - all terms which are important in physics. If I were younger, I would attack Noether over p-adics. Who knows, it might be a new unknown way to describe motion, and therewith physics. It isn't crazier than to search for bosinos. The main obstacle is probably not the analysis or at least differentiation, it is likely to develop an intuition for those fields.

WWGD

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I always found quadratic forms with Q(a,a)=0 for nonzero a intriguing. Positive-definite ones and inner- products somehow feel more reasonable.

fresh_42

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Wikipedia says something about p-adic analysis, and differentiation!
$$f\, : \,\mathbb{Q}_p \longrightarrow \mathbb{Q}_p \; , \;x \longmapsto \left(\dfrac{1}{|x|_p}\right)^2 \text{ and } f(0)=0$$
is differentiable on $\mathbb{Q}_p$ with $f\,'(x)\equiv 0$, but it isn't even locally constant in $x=0$.

This means a differentiable p-adic manifold is fundamentally different than a real one.

fresh_42

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I think you may be able to do it over the p-adics, but I have no idea how/what that would be like.
Btw., do you have any imagination how an open ball in a p-adic topology looks like?

WWGD

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Btw., do you have any imagination how an open ball in a p-adic topology looks like?
Not now, let me think it through.

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